试题

题目:
计算:
(1)
12
-
48
+
32

(2)
45
÷
1
5
×
2
2
3

(3)
3a2
÷3
a
2
×
1
2
2a
3

(4)(
3
-
2
)2×(5+2
6
)

答案
解:(1)
12
-
48
+
32

=2
3
-4
3
+4
2

=(2-4)
3
+4
2

=-2
3
+4
2


(2)
45
÷
1
5
×
2
2
3

=
45
×
5
×
2
2
3

=
45×5×
8
3

=10
6


(3)
3a2
÷3
a
2
×
1
2
2a
3

=
3a 2
×
1
3
2
a
×
1
2
2a
3

=
1
3
×
1
2
×
3a 2×
2
a
×
2a
3

=
1
6
×2a,
=
a
3


(4)(
3
-
2
)2×(5+2
6
)

=(3+2-2
6
)×(5+2
6
),
=(5-2
6
)×(5+2
6
),
=25-24,
=1.
解:(1)
12
-
48
+
32

=2
3
-4
3
+4
2

=(2-4)
3
+4
2

=-2
3
+4
2


(2)
45
÷
1
5
×
2
2
3

=
45
×
5
×
2
2
3

=
45×5×
8
3

=10
6


(3)
3a2
÷3
a
2
×
1
2
2a
3

=
3a 2
×
1
3
2
a
×
1
2
2a
3

=
1
3
×
1
2
×
3a 2×
2
a
×
2a
3

=
1
6
×2a,
=
a
3


(4)(
3
-
2
)2×(5+2
6
)

=(3+2-2
6
)×(5+2
6
),
=(5-2
6
)×(5+2
6
),
=25-24,
=1.
考点梳理
二次根式的混合运算.
(1)利用二次根式的运算性质,加减运算首先进行化简开平方,使二次根式达到最简,再进行合并被开方数相同的二次根式;
(2)利用二次根式的运算性质,除法首先变成乘法运算,根号内的与根号内的相乘,根号外的与根号外的相乘;
(3)利用以上性质进行运算;
(4)利用以上性质进行运算;
此题主要考查了二次根式的混合运算,在运算二次根式的乘除运算时,根号内的与根号内的相乘,根号外的与根号外的相乘;
是解决问题的关键.
计算题.
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